Difference between revisions of "1997 AIME Problems/Problem 8"
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===Solution 2=== | ===Solution 2=== | ||
Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of <math>6</math> ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange one 1 and 2 -1's. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is <math>6*3*(2+1+2)</math> = <math> \boxed{090}</math>. | Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of <math>6</math> ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange one 1 and 2 -1's. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is <math>6*3*(2+1+2)</math> = <math> \boxed{090}</math>. | ||
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== See also == | == See also == | ||
Revision as of 01:54, 2 January 2020
Problem
How many different 
arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
Solution
Solution 1
- For more detailed explanations, see related problem (AIME I 2007, 10).
The problem is asking us for all configurations of grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:
- The first two columns share no two numbers in the same row. There are
ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are 
ways. This gives 
. - The first two columns share one number in the same row. There are
ways to pick the position of the shared 1, then 
ways to pick the locations for the next two 1s, and then 
ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in ways. This gives 
. - The first two columns share two numbers in the same row. There are
ways to pick the position of the shared 1s. Everything is then fixed.
ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are 
ways. This gives 
.
ways to pick the position of the shared 1, then 
ways to pick the locations for the next two 1s, and then 
ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in 
.
ways to pick the position of the shared 1s. Everything is then fixed.Adding these cases up, we get 
.
Solution 2
Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of 
ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange one 1 and 2 -1's. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is 
= 
.
-pi_is_3.141
See also
- 2007 AIME I Problems/Problem 10 - the same problem, but with a
| 1997 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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